Question

Temperature of diatomic gas is 360 K. If moment of inertia of its molecules is $8.28\times {10}^{-38}g-c{m}^2$. Calculate their root mean square angular velocity.

Answer 1

Since moment of inertia is given by $I=M{R}^2$

Also, $v_{rms}=\sqrt{\frac{3RT}{M}}$

Hence, ${\omega }_{rms}=\frac{v_{rms}}{R}=\sqrt{\frac{3RT}{M{R}^2}}=\sqrt{\frac{3RT}{I}}$

${\omega }_{rms}=\sqrt{\frac{3\left(8.314\right)\left(360\right)}{8.28X{10}^{-42}}}=\sqrt{1.02X{10}^{26}}\approx {10}^{13}rad/s$

Answer is ${10}^{13}rad/s$